Prove $\sqrt{2} + \sqrt{5}$ is irrational

Question

How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?

I tried to prove it by contradiction and got this equation: $a^2/b^2 = \sqrt{40}$.

See here http://math.stackexchange.com/questions/496037/prove-that-sqrt-2-sqrt-3-is-irrational/496102#496102 — Supriyo, Oct 25 '13 at 13:08
I have posted answer for this two times in this forum... please try checking with some key words... — , Oct 25 '13 at 13:09
Square the number: if it's rational, then also $\sqrt{10}$ is. Is it? — egreg, Oct 25 '13 at 13:10
@PraphullaKoushik Which key words? $\sqrt{2}$ isn't exactly eacy to search for... — user1729, Oct 25 '13 at 13:55

score 24 · Answer 1 · answered Oct 25 '13 at 16:18

24

$\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 5-\dfrac 3{\sqrt 2+\sqrt 5}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 5\notin \mathbb Q$

answered Oct 25 '13 at 16:18

Philippe Malot

2,667

1233dfv · Accepted Answer · 2013-10-25T14:45:30.550

Use proof by contradiction. Assume that the sum is rationial, that is $$\sqrt2 +\sqrt5 = {a\over b}$$ where $a$ and $b$ are integers with $b\neq0$. Now rewrite this as $$\sqrt5={a\over b}-\sqrt2.$$ Squaring both sides of this equation we obtain $$5={a^2\over b^2}-2\sqrt2{a\over b}+2.$$ Now, carefully solve for $\sqrt2$ and obtain $$\sqrt2={-3b\over 2a}+{a\over 2b.}$$ This implies that $\sqrt2$ is a rational number which is a contradiction. Thus $$\sqrt2+\sqrt5$$ is an irrational number.

ncmathsadist · Answer 3 · 2013-10-25T19:16:38.353

12

You have $$(\sqrt{2} + \sqrt{5})^2 = 7 + 2\sqrt{10}.$$
The square of a rational number is rational. This number, $7 + 2\sqrt{10}$, is rational iff $\sqrt{10}$ is rational. The standard argument shows that $\sqrt{10}$ is not rational. So we are done here.

edited Oct 25 '13 at 19:16

answered Oct 25 '13 at 14:55

ncmathsadist

49,383

You're almost correct. Square of a rational number is rational, but the square of an irrational number can be rational. So the "iff" is wrong in this context, but one implication, which is the one we use, is in fact true. – Asaf Karagila Oct 25 '13 at 15:08
I am saying $9 + 2\sqrt{10}$ is rational iff $\sqrt{10}$ is ratioinal. – ncmathsadist Oct 25 '13 at 15:54
Thanks for the comment, @Thomas – ncmathsadist Oct 25 '13 at 19:17
Ah, that's better. – Asaf Karagila Oct 25 '13 at 19:38

score 9 · Answer 4 · answered Oct 25 '13 at 14:58

9

Let $\sqrt2+\sqrt5=a$ where $a$ is rational

$\implies\sqrt2=a-\sqrt5$

Squaring we get, $$2=a^2+5-2a\sqrt5\iff\sqrt5=\frac{a^2+3}{2a}$$ which is rational unlike $\sqrt5$

answered Oct 25 '13 at 14:58

lab bhattacharjee

274,582

can you explain how you went from the squaring to this result 2=a2+5−2a5–√5 – A P Mar 08 '22 at 13:14
1

@AviPars, https://en.wikipedia.org/wiki/Proof_by_contradiction#Irrationality_of_the_square_root_of_2 – lab bhattacharjee Mar 08 '22 at 13:33

score 7 · Answer 5 · answered Oct 25 '13 at 13:50

If your arrival at the equation $a^2/b^2=\sqrt{40}$ was correct, you’re done, because the equation $a^4=40b^4$ for integers $a$ and $b$ is a contradiction to the Fundamental Theorem of Arithmetic, which says that the expression of an integer as product of primes can be done in only one way. But your suspect fourth-degree equation has a number of $5$’s on the left that is divisible by $4$, but not so on the right.

Elias Costa · Answer 6 · 2013-10-25T20:06:24.510

If $\sqrt{2}+\sqrt{5}\in\mathbb{Q}$, i.e., $ \sqrt{2}+\sqrt{5}=\frac{b}{a}$ for $a,b\in\mathbb{Z} \mbox{ and } a\neq 0$ then $ 2-5=\frac{b}{a}(\sqrt{2}-\sqrt{5})$ and $$ \sqrt{2}-\sqrt{5} =\frac{-3\cdot a}{b} $$ Here $b$ is necessarily greater than zero by cause $\sqrt{2}+\sqrt{5}>0$.Finally we get, $$ \sqrt{2}=\frac{1}{2}\left( (\sqrt{2}+\sqrt{5})+(\sqrt{2}-\sqrt{5}) \right)= \frac{1}{2}\left( \frac{b}{a}+\frac{-3\cdot a}{b} \right)\in\mathbb{Q}. $$ But this is a contradiction since $\sqrt{2}$ is not rational.

Prove $\sqrt{2} + \sqrt{5}$ is irrational

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